已知tan a=2 ,2sin a cos a=?(sina)^2-sin a cos a -2(cos a)^2=?
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已知tan a=2 ,2sin a cos a=?(sina)^2-sin a cos a -2(cos a)^2=?
![已知tan a=2 ,2sin a cos a=?(sina)^2-sin a cos a -2(cos a)^2=?](/uploads/image/z/8003073-57-3.jpg?t=%E5%B7%B2%E7%9F%A5tan+a%3D2+%2C2sin+a+cos+a%3D%3F%28sina%29%5E2-sin+a+cos+a+-2%28cos+a%29%5E2%3D%3F)
利用sin^2 a+cos^2 a=1,
2sin a cos a
=2sin a cos a/(sin^2 a+cos^2 a)
=2tana/(tan^2 a+1)
=2x2/(4+1)
=4/5.
(sina)^2-sin a cos a -2(cos a)^2
=[(sina)^2-sin a cos a -2(cos a)^2]/(sin^2 a+cos^2 a)
=(tan^2 a-tan a -2)/(tan^2 a+1)
=(4-2-2)/(4+1)
=0.
2sin a cos a
=2sin a cos a/(sin^2 a+cos^2 a)
=2tana/(tan^2 a+1)
=2x2/(4+1)
=4/5.
(sina)^2-sin a cos a -2(cos a)^2
=[(sina)^2-sin a cos a -2(cos a)^2]/(sin^2 a+cos^2 a)
=(tan^2 a-tan a -2)/(tan^2 a+1)
=(4-2-2)/(4+1)
=0.
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